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Ch.20 - Electrochemistry
All textbooksBrown 15th EditionCh.20 - ElectrochemistryProblem 73
Chapter 20, Problem 73

A voltaic cell is constructed with two Zn²⁺/Zn electrodes. The two half-cells have [Zn²⁺] = 1.8 M and [Zn²⁺] = 1.00 × 10⁻² M, respectively. (a) Which electrode is the anode of the cell? (b) What is the standard emf of the cell? (c) What is the cell emf for the concentrations given? (d) For each electrode, predict whether [Zn²⁺] will increase, decrease, or stay the same as the cell operates.

Verified step by step guidance
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Step 1: Identify the anode and cathode by comparing the concentrations of Zn²⁺ ions. The anode is where oxidation occurs, and the cathode is where reduction occurs. In a concentration cell, the anode is the electrode with the lower concentration of Zn²⁺ ions.
Step 2: Determine the standard emf (E°) of the cell. For a concentration cell with identical electrodes, the standard emf is 0 V because the electrodes are the same material.
Step 3: Use the Nernst equation to calculate the cell emf for the given concentrations: E = E° - (RT/nF) * ln(Q), where Q is the reaction quotient, R is the gas constant, T is the temperature in Kelvin, n is the number of moles of electrons transferred, and F is Faraday's constant.
Step 4: Calculate the reaction quotient Q for the cell: Q = [Zn²⁺] at anode / [Zn²⁺] at cathode. Substitute the given concentrations into the equation.
Step 5: Predict the change in [Zn²⁺] for each electrode as the cell operates. At the anode, [Zn²⁺] will increase as Zn metal oxidizes to Zn²⁺ ions. At the cathode, [Zn²⁺] will decrease as Zn²⁺ ions are reduced to Zn metal.

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Voltaic Cell and Electrodes

A voltaic cell, also known as a galvanic cell, generates electrical energy through spontaneous redox reactions. It consists of two electrodes: the anode, where oxidation occurs, and the cathode, where reduction takes place. In this context, the electrode with the higher concentration of Zn²⁺ ions will act as the cathode, while the one with the lower concentration will be the anode.
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Nernst Equation

The Nernst equation relates the cell potential (emf) to the concentrations of the reactants and products in a redox reaction. It allows for the calculation of the cell emf under non-standard conditions by incorporating the reaction quotient. This equation is crucial for determining how the emf changes with varying concentrations of Zn²⁺ in the half-cells.
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Concentration Changes in Electrodes

As a voltaic cell operates, the concentrations of the reactants and products change. At the anode, oxidation leads to a decrease in the concentration of Zn²⁺ ions, while at the cathode, reduction results in an increase in Zn²⁺ concentration. Understanding these changes is essential for predicting the behavior of the cell as it discharges.
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Related Practice
Textbook Question

A voltaic cell utilizes the following reaction: 4 Fe2+1aq2 + O21g2 + 4 H+1aq2 ¡ 4 Fe3+1aq2 + 2 H2O1l2 (b) What is the emf of this cell when 3Fe2+4 = 1.3 M, 3Fe3+4= 0.010 M, PO2 = 0.50 atm, and the pH of the solution in the cathode half-cell is 3.50?

Textbook Question

A voltaic cell utilizes the following reaction: 2 Fe3+1aq2 + H21g2 ¡ 2 Fe2+1aq2 + 2 H+1aq2 (a) What is the emf of this cell under standard conditions?

Textbook Question

A voltaic cell utilizes the following reaction: (b) What is the emf for this cell when 3Fe3+4 = 3.50 M, PH2= 0.95 atm, 3Fe2+4 = 0.0010 M, and the pH in both half-cells is 4.00?

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Textbook Question

A voltaic cell is constructed that is based on the following reaction: Sn2+(aq) + Pb(s) → Sn(s) + Pb2+(aq) (a) If the concentration of Sn2+ in the cathode half-cell is 1.00 M and the cell generates an emf of +0.22 V, what is the concentration of Pb2+ in the anode half-cell?

Textbook Question

During a period of discharge of a lead–acid battery, 402 g of Pb from the anode is converted into PbSO4(s). (a) What mass of PbO2(s) is reduced at the cathode during this same period?

Textbook Question

During a period of discharge of a lead–acid battery, 402 g of Pb from the anode is converted into PbSO4(s). (b) How many coulombs of electrical charge are transferred from Pb to PbO2?